{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Aersp_311_Exam_I_08 - Aerospace Engineering 311 Exam I...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Aerospace Engineering 311 Exam I 9/25/08 Problem 1 — Multiple Choicegst (5 Points) Select the most accurate ending(s). The intermolecular force that allows spiders to walk on water, a steel needle to float when placed gently in a pan of water and water “heads” on your freshly waxed car is known as b) Newtonian c) vorticity d) viscosity e) density 0 Brungart Problem 2 - Short Answer (10 Points) Explain the basic difierences between the Eulerian and Lagrangian description of fluid flow. BAUER. 366W3P‘L‘lom— Ob6¢W65 Hue. Mo'LlaDI‘b 3+0 DEE¢L£JPaPéILI¢5 ad: H16?! MOVE. fibrcugh a at: HELA 6Pam. or- PO ' rub» (5 Pomfil . ' __ w ' b I 5;}; H36 heft. a. Lagranalari. clasu‘tp‘lzlom .flngws iron»; cilia: abilwegand chasm becl ab 14: Moves gpaca- (5 Petra?) Problem 3 [11 Points) An aircraft flies at a velocity of 67 m/s where the temperature is 15.6“C and the pressure is 1 atm. (a) What is the Mach number of the flow? (5 Points) M = 3% <3 Pom) Davin. c.10- mam. _. VIM/5 / H - :: am, 2 P .4145) germ/5 “ o (b) Assuming that the Mach number is 0.56, is the flow about the aircraft considered to be compressible or incompressible? Why? (6 Points) Emgmz-c. M 5:03 - InCDMPFBGSIIDIE- (4 pomfi) M >05 - wnpresslbie. (H: 0-56) 7 0-3 s" COHPmsibiL (2 Ronnie») Problem 4 (9 Points) A barrel 1 m in diameter and containing an unknown liquid 2.0 m deep is exposed to that atmosphere at sea level. The pressure at the bottom of the barrel is 125,721 Pa. Determine the density of the unknown liquid. Hgérob-iza-L'Ic. rials-1:101. p= Pa - 63 E ('5 Fin-Lian) - o=-€31 %=I0li¢P& P P P F __ __ o- e - a? Zn : M ( F 4,43 9 (9.6062) ‘3 ‘3' i =12572IP2. a= 1.2m lash-t3 (:Po.q4:\ Problem 5 [40 Points Total} A layer of water flows down an. inclined fixed surface with the velocity profile shown below. (a) Determine the magnitude and direction of the shearing stress that the water exerts on the fixed Surface for U = 3 mis and h = 0.1 m. (10 Points) 7;: (2 Ennis) /Z:N__: floured. 7:0 5 IN: ace “m (I at.) In cin: o-izflow (I Pt.) (b) Determine the mass flux through a plane normal to the fixed surface for U = 3 mls and h = 0.1 m. Assume that the flow extends a distance of l to into the page (i.e., z-dir.). (15 Points) 9-,: Jam- U%)A% (333.4145) Ewe—WI; f1: l‘i‘tbksla (l fining Problem 5 Continued (40 Point‘s Total! (c) Determine the momentum flux through a plane normal to the fixed surface for U = 3 nuts and h = 0.1 m. Assume that the flow extends a distance of 1 m into the page (i.e., z-dir.). (15 Points) ‘15:: {Lam-mas (1 mt) h i z z 4 a. ‘6 ““‘§%§— 4% +4%% 8 Pow—*3) Z : 0.533631% §$K5M4f<Q1MX|Q l; “1: km N: Lam = 478.719 (1 Bauer) § 5 52 25 Problem 5( Points Total! Expand g:- into the x, y and z coordinate directions for a general velocity field. In other words, determine the acceleration vector for the general velocity field given below. ii = 110:, y, z, 01' + v(x. y, z, 0}" + w(x, y, z, 075 D; _ “ a —- -=~ Bil-— fin (u-vw (lRmn-E) 6 =A§+fi+k§z (1'6: (Lu-L‘fl‘r W12.) (£3? “51% + DIE cm * _.... T=E£+CU°§7W :AEE+U£+\J%+W%1+AEE+U%+V£+w%1 (Z Bowl?) 40" cal-rem! J—flrm.) Useful Material For Exam 1 g m pdV 41,96 ofidS = 0 S g = 9.81 We2 13: jpmz? ofi)dS S 'l'lhln A3 hope-Lina of a . . m m M Cm Liquidsall film that! gig/If mtg/(mm) Y.N/nl‘ p, N/In’ N/m‘ pal-m0 “mm” mm as my 2.13 3-2 mom-s —. ms Bum: 881 HI 3-4 1383-1 “3134-4 1.4 3+9 4.34 mm 1.990 9.6734 1733-3 LEE-Ht 955 3+! 4.45 My. . __ -- __ __ m _ 1103-3 2233-2 5.? 3+3 9.0 3+! 5.1! Elma: 5mm __l.llf{_u_114 up; w 4343-2 1.: 5+1 - 11.1 Finn 12 1.337 1623-4 — -— -- 1.16 (hauling 680 an 3-4 2.16 3-2 5.51344 9.53 3+8 3.68 (“slain - 1.2m m 6.33 5-2 :4 3-2 4343+, 23.0 m 50‘ 1.92 3-3 13 34 3." 3+3 1.6 3+9 5.56 — My ' ‘ '7 ‘- 13.590 :1.“ 3-3 43‘ 3-] M 3-3 2.55 3+“) 1.0? Mahmud 791 5383-4 2253-2 l343+4 8.3 3+8 4.63 SAE 1W all 810 m4 B-l‘ 3.6 3-2 — 13134-9 15.“! 8A3 10W all 876 1.7 3-1‘ — — — 14.0 SAEIIWGII _ _ $1 2.9 51* 3.5 3-2 — 1.38 3+9 18.3 SABiGWdl 9a: 3.6 34‘ — - — 20.: Vista 998 1.1:: 3-3 723 3-2 2343+! 2.19 3+9 TINSAJ Sammy) 1M 1313-3 7.” 3-2 2343+! 2333+9 7.28 ...
View Full Document

Report

{[ snackBarMessage ]}